If a `prop ` b and `prop ` c , then show that `(a^3b^3+b^3c^3+c^3a^3) prop abc (a^3+b^3+c^3)`.

Explanation

`a prop b rArr a =k_(1)b (k_(1)` =non -zero variation constant )
`b prop c rArr b= k_(2)c (k_(2)`=non -zero variation constant )
`therefore a= k_(1) k_(2) c ` .
Now , `(a^3b^3+b^3c^3+c^3a^3)/(abc(a^3+b^3+c^3))=((k_(1)k_2c)^3+(k_2c)^3+(k_2c)^3+(c)^3+c^3.(k_1k_2c)^3)/(k_1k_2c.k_2c.c{(k_1k_2c)^3(k_2c)^3+c^3}`
`(k_1^3k_2^6c^6+k_2^3c^6+k_1^3k_2^3c^6)/(k_1k_2^2c^3(k_1^3k_2^3c^3+k_2^3c^3+c^3))`
`=(k_2^3c^6(k_1^3k_2^3+1+k_1^3))/(k_1k_2^2c^6(k_1^3k_2^3+k_2^3+1))`
`=(k_2(k_1^3k_2^3+k_1^3+1))/(k_1(k_1^3k_2^3+k_2^3+1))`
=Non -zero constant `[because k-1 and k_2 ne0]`
=k (let)
`therefore a^3b^3+b^3c^3+c^3a^3=k{abc(a^3+b^3+c^3)}`
`therefore a^3b^3+b^3c^3+c^3a^3 prop abc (a^3+b^3+c^3)` [`because k ne0=` variation constant. ] (proved )